将所有 1 组合在一起所需的最小互换量
原文:https://www . geesforgeks . org/minimum-swaps-required-group-1s-together/
给定一个 0 和 1 的数组，我们需要编写一个程序来找到将数组中的所有 1 组合在一起所需的最小交换次数。
示例:
Input : arr[] = {1, 0, 1, 0, 1}
Output : 1
Explanation: Only 1 swap is required to
group all 1's together. Swapping index 1
and 4 will give arr[] = {1, 1, 1, 0, 0}
Input : arr[] = {1, 0, 1, 0, 1, 1}
Output : 1
一个简单的解决方法是首先计算数组中 1 的总数。假设这个计数是 x，现在我们需要找到这个数组的长度为 x 的子数组，最大数量为 1。所需的最小交换将是长度为 x 的子数组中 0 的数量，最大数量为 1。 时间复杂度:O(n 2
一个有效的解决方案是使用滑动窗口技术的概念优化上述方法中寻找子阵列的蛮力技术。我们可以维持一个预计数阵列，以寻找在 0(1)时间复杂度的子阵列中存在的 1 的数量。
以下是上述想法的实现:
C++
// C++ program to find minimum swaps
// required to group all 1's together
#include <iostream>
#include <limits.h>
using namespace std;
// Function to find minimum swaps
// to group all 1's together
int minSwaps(int arr[], int n) {
int noOfOnes = 0;
// find total number of all in the array
for (int i = 0; i < n; i++) {
if (arr[i] == 1)
noOfOnes++;
}
// length of subarray to check for
int x = noOfOnes;
int maxOnes = INT_MIN;
// array to store number of 1's upto
// ith index
int preCompute[n] = {0};
// calculate number of 1's upto ith
// index and store in the array preCompute[]
if (arr[0] == 1)
preCompute[0] = 1;
for (int i = 1; i < n; i++) {
if (arr[i] == 1) {
preCompute[i] = preCompute[i - 1] + 1;
} else
preCompute[i] = preCompute[i - 1];
}
// using sliding window technique to find
// max number of ones in subarray of length x
for (int i = x - 1; i < n; i++) {
if (i == (x - 1))
noOfOnes = preCompute[i];
else
noOfOnes = preCompute[i] - preCompute[i - x];
if (maxOnes < noOfOnes)
maxOnes = noOfOnes;
}
// calculate number of zeros in subarray
// of length x with maximum number of 1's
int noOfZeroes = x - maxOnes;
return noOfZeroes;
}
// Driver Code
int main() {
int a[] = {1, 0, 1, 0, 1};
int n = sizeof(a) / sizeof(a[0]);
cout << minSwaps(a, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find minimum swaps
// required to group all 1's together
import java.io.*;
class GFG {
// Function to find minimum swaps
// to group all 1's together
static int minSwaps(int arr[], int n) {
int noOfOnes = 0;
// find total number of all in the array
for (int i = 0; i < n; i++) {
if (arr[i] == 1)
noOfOnes++;
}
// length of subarray to check for
int x = noOfOnes;
int maxOnes = Integer.MIN_VALUE;
// array to store number of 1's upto
// ith index
int preCompute[] = new int[n];
// calculate number of 1's upto ith
// index and store in the array preCompute[]
if (arr[0] == 1)
preCompute[0] = 1;
for (int i = 1; i < n; i++) {
if (arr[i] == 1) {
preCompute[i] = preCompute[i - 1] + 1;
} else
preCompute[i] = preCompute[i - 1];
}
// using sliding window technique to find
// max number of ones in subarray of length x
for (int i = x - 1; i < n; i++) {
if (i == (x - 1))
noOfOnes = preCompute[i];
else
noOfOnes = preCompute[i] - preCompute[i - x];
if (maxOnes < noOfOnes)
maxOnes = noOfOnes;
}
// calculate number of zeros in subarray
// of length x with maximum number of 1's
int noOfZeroes = x - maxOnes;
return noOfZeroes;
}
// Driver Code
public static void main (String[] args) {
int a[] = {1, 0, 1, 0, 1};
int n = a.length;
System.out.println( minSwaps(a, n));
}
}
// This code is contributed by vt_m.
Python 3
# Python program to
# find minimum swaps
# required to group
# all 1's together
# Function to find minimum swaps
# to group all 1's together
def minSwaps(arr,n):
noOfOnes = 0
# find total number of
# all in the array
for i in range(n):
if (arr[i] == 1):
noOfOnes=noOfOnes+1
# length of subarray to check for
x = noOfOnes
maxOnes = -2147483648
# array to store number of 1's upto
# ith index
preCompute={}
# calculate number of 1's upto ith
# index and store in the
# array preCompute[]
if (arr[0] == 1):
preCompute[0] = 1
for i in range(1,n):
if (arr[i] == 1):
preCompute[i] = preCompute[i - 1] + 1
else:
preCompute[i] = preCompute[i - 1]
# using sliding window
# technique to find
# max number of ones in
# subarray of length x
for i in range(x-1,n):
if (i == (x - 1)):
noOfOnes = preCompute[i]
else:
noOfOnes = preCompute[i] - preCompute[i - x]
if (maxOnes < noOfOnes):
maxOnes = noOfOnes
# calculate number of zeros in subarray
# of length x with maximum number of 1's
noOfZeroes = x - maxOnes
return noOfZeroes
# Driver code
a = [1, 0, 1, 0, 1]
n = len(a)
print(minSwaps(a, n))
# This code is contributed
# by Anant Agarwal.
C
// C# program to find minimum swaps
// required to group all 1's together
using System;
class GFG {
// Function to find minimum swaps
// to group all 1's together
static int minSwaps(int []arr, int n) {
int noOfOnes = 0;
// find total number of all in the array
for (int i = 0; i < n; i++) {
if (arr[i] == 1)
noOfOnes++;
}
// length of subarray to check for
int x = noOfOnes;
int maxOnes = int.MinValue;
// array to store number of 1's upto
// ith index
int []preCompute = new int[n];
// calculate number of 1's upto ith
// index and store in the array preCompute[]
if (arr[0] == 1)
preCompute[0] = 1;
for (int i = 1; i < n; i++) {
if (arr[i] == 1) {
preCompute[i] = preCompute[i - 1] + 1;
} else
preCompute[i] = preCompute[i - 1];
}
// using sliding window technique to find
// max number of ones in subarray of length x
for (int i = x - 1; i < n; i++) {
if (i == (x - 1))
noOfOnes = preCompute[i];
else
noOfOnes = preCompute[i] - preCompute[i - x];
if (maxOnes < noOfOnes)
maxOnes = noOfOnes;
}
// calculate number of zeros in subarray
// of length x with maximum number of 1's
int noOfZeroes = x - maxOnes;
return noOfZeroes;
}
// Driver Code
public static void Main ()
{
int []a = {1, 0, 1, 0, 1};
int n = a.Length;
Console.WriteLine( minSwaps(a, n));
}
}
// This code is contributed by vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find minimum swaps
// required to group all 1's together
// Function to find minimum swaps
// to group all 1's together
function minSwaps($arr, $n)
{
$noOfOnes = 0;
// find total number of
// all in the array
for($i = 0; $i < $n; $i++)
{
if ($arr[$i] == 1)
$noOfOnes++;
}
// length of subarray
// to check for
$x = $noOfOnes;
$maxOnes = PHP_INT_MIN;
// array to store number of
// 1's upto ith index
$preCompute = array();
// calculate number of 1's
// upto ith index and store
// in the array preCompute[]
if ($arr[0] == 1)
$preCompute[0] = 1;
for($i = 1; $i < $n; $i++)
{
if ($arr[$i] == 1)
{
$preCompute[$i] = $preCompute[$i - 1] + 1;
}
else
$preCompute[$i] = $preCompute[$i - 1];
}
// using sliding window
// technique to find
// max number of ones in
// subarray of length x
for ( $i = $x - 1; $i < $n; $i++)
{
if ($i == ($x - 1))
$noOfOnes = $preCompute[$i];
else
$noOfOnes = $preCompute[$i] -
$preCompute[$i - $x];
if ($maxOnes < $noOfOnes)
$maxOnes = $noOfOnes;
}
// calculate number of zeros in subarray
// of length x with maximum number of 1's
$noOfZeroes = $x - $maxOnes;
return $noOfZeroes;
}
// Driver Code
$a = array(1, 0, 1, 0, 10);
$n = count($a);
echo minSwaps($a, $n);
// This code is contributed by anuj_67.
?>
java 描述语言
<script>
// Javascript program to find minimum swaps
// required to group all 1's together
// Function to find minimum swaps
// to group all 1's together
function minSwaps(arr, n) {
let noOfOnes = 0;
// find total number of all in the array
for (let i = 0; i < n; i++) {
if (arr[i] == 1)
noOfOnes++;
}
// length of subarray to check for
let x = noOfOnes;
let maxOnes = Number.MIN_VALUE;
// array to store number of 1's upto
// ith index
let preCompute = new Array(n);
preCompute.fill(0);
// calculate number of 1's upto ith
// index and store in the array preCompute[]
if (arr[0] == 1)
preCompute[0] = 1;
for (let i = 1; i < n; i++) {
if (arr[i] == 1) {
preCompute[i] = preCompute[i - 1] + 1;
} else
preCompute[i] = preCompute[i - 1];
}
// using sliding window technique to find
// max number of ones in subarray of length x
for (let i = x - 1; i < n; i++) {
if (i == (x - 1))
noOfOnes = preCompute[i];
else
noOfOnes = preCompute[i] - preCompute[i - x];
if (maxOnes < noOfOnes)
maxOnes = noOfOnes;
}
// calculate number of zeros in subarray
// of length x with maximum number of 1's
let noOfZeroes = x - maxOnes;
return noOfZeroes;
}
let a = [1, 0, 1, 0, 1];
let n = a.length;
document.write( minSwaps(a, n));
</script>
Output
1
时间复杂度:O(n) T3】辅助空间 : O(n)
另一种有效的方法: 首先统计数组中 1 的总数。假设这个计数是 x，现在使用窗口滑动技术的概念找到这个最大数量为 1 的数组的长度为 x 的子数组。维护一个变量，以便在 0(1)个额外空间中找到子数组中存在的 1 的数量，并且为每个子数组维护最大 1 变量，最后返回 0 的数量(0 的数量= x–最大 1)。
C++
// C++ code for minimum swaps
// required to group all 1's together
#include <iostream>
#include <limits.h>
using namespace std;
// Function to find minimum swaps
// to group all 1's together
int minSwaps(int arr[], int n)
{
int numberOfOnes = 0;
// find total number of all 1's in the array
for (int i = 0; i < n; i++) {
if (arr[i] == 1)
numberOfOnes++;
}
// length of subarray to check for
int x = numberOfOnes;
int count_ones = 0, maxOnes;
// Find 1's for first subarray of length x
for(int i = 0; i < x; i++){
if(arr[i] == 1)
count_ones++;
}
maxOnes = count_ones;
// using sliding window technique to find
// max number of ones in subarray of length x
for (int i = 1; i <= n-x; i++) {
// first remove leading element and check
// if it is equal to 1 then decrement the
// value of count_ones by 1
if (arr[i-1] == 1)
count_ones--;
// Now add trailing element and check
// if it is equal to 1 Then increment
// the value of count_ones by 1
if(arr[i+x-1] == 1)
count_ones++;
if (maxOnes < count_ones)
maxOnes = count_ones;
}
// calculate number of zeros in subarray
// of length x with maximum number of 1's
int numberOfZeroes = x - maxOnes;
return numberOfZeroes;
}
// Driver Code
int main() {
int a[] = {0, 0, 1, 0, 1, 1, 0, 0, 1};
int n = sizeof(a) / sizeof(a[0]);
cout << minSwaps(a, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// java program to find largest number
// smaller than equal to n with m set
// bits then m-1 0 bits.
public class GFG {
// Function to find minimum swaps
// to group all 1's together
static int minSwaps(int arr[], int n)
{
int numberOfOnes = 0;
// find total number of all 1's
// in the array
for (int i = 0; i < n; i++) {
if (arr[i] == 1)
numberOfOnes++;
}
// length of subarray to check for
int x = numberOfOnes;
int count_ones = 0, maxOnes;
// Find 1's for first subarray
// of length x
for(int i = 0; i < x; i++){
if(arr[i] == 1)
count_ones++;
}
maxOnes = count_ones;
// using sliding window technique
// to find max number of ones in
// subarray of length x
for (int i = 1; i <= n-x; i++) {
// first remove leading element
// and check if it is equal to
// 1 then decrement the
// value of count_ones by 1
if (arr[i - 1] == 1)
count_ones--;
// Now add trailing element
// and check if it is equal
// to 1 Then increment the
// value of count_ones by 1
if(arr[i + x - 1] == 1)
count_ones++;
if (maxOnes < count_ones)
maxOnes = count_ones;
}
// calculate number of zeros in
// subarray of length x with
// maximum number of 1's
int numberOfZeroes = x - maxOnes;
return numberOfZeroes;
}
// Driver code
public static void main(String args[])
{
int a[] = new int[]{0, 0, 1, 0,
1, 1, 0, 0, 1};
int n = a.length;
System.out.println(minSwaps(a, n));
}
}
// This code is contributed by Sam007
Python 3
# Python code for minimum
# swaps required to group
# all 1's together
# Function to find minimum
# swaps to group all 1's
# together
def minSwaps(arr, n) :
numberOfOnes = 0
# find total number of
# all 1's in the array
for i in range(0, n) :
if (arr[i] == 1) :
numberOfOnes = numberOfOnes + 1
# length of subarray
# to check for
x = numberOfOnes
count_ones = 0
maxOnes = 0
# Find 1's for first
# subarray of length x
for i in range(0, x) :
if(arr[i] == 1) :
count_ones = count_ones + 1
maxOnes = count_ones
# using sliding window
# technique to find
# max number of ones in
# subarray of length x
for i in range(1, (n - x + 1)) :
# first remove leading
# element and check
# if it is equal to 1
# then decrement the
# value of count_ones by 1
if (arr[i - 1] == 1) :
count_ones = count_ones - 1
# Now add trailing
# element and check
# if it is equal to 1
# Then increment
# the value of count_ones by 1
if(arr[i + x - 1] == 1) :
count_ones = count_ones + 1
if (maxOnes < count_ones) :
maxOnes = count_ones
# calculate number of
# zeros in subarray
# of length x with
# maximum number of 1's
numberOfZeroes = x - maxOnes
return numberOfZeroes
# Driver Code
a = [0, 0, 1, 0, 1, 1, 0, 0, 1]
n = 9
print (minSwaps(a, n))
# This code is contributed
# by Manish Shaw(manishshaw1)
C
// C# code for minimum swaps
// required to group all 1's together
using System;
class GFG{
// Function to find minimum swaps
// to group all 1's together
static int minSwaps(int []arr, int n)
{
int numberOfOnes = 0;
// find total number of all 1's in the array
for (int i = 0; i < n; i++) {
if (arr[i] == 1)
numberOfOnes++;
}
// length of subarray to check for
int x = numberOfOnes;
int count_ones = 0, maxOnes;
// Find 1's for first subarray of length x
for(int i = 0; i < x; i++){
if(arr[i] == 1)
count_ones++;
}
maxOnes = count_ones;
// using sliding window technique to find
// max number of ones in subarray of length x
for (int i = 1; i <= n-x; i++) {
// first remove leading element and check
// if it is equal to 1 then decrement the
// value of count_ones by 1
if (arr[i - 1] == 1)
count_ones--;
// Now add trailing element and check
// if it is equal to 1 Then increment
// the value of count_ones by 1
if(arr[i + x - 1] == 1)
count_ones++;
if (maxOnes < count_ones)
maxOnes = count_ones;
}
// calculate number of zeros in subarray
// of length x with maximum number of 1's
int numberOfZeroes = x - maxOnes;
return numberOfZeroes;
}
// Driver Code
static public void Main ()
{
int []a = {0, 0, 1, 0, 1, 1, 0, 0, 1};
int n = a.Length;
Console.WriteLine(minSwaps(a, n));
}
}
// This code is contributed by vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP code for minimum swaps
// required to group all 1's together
// Function to find minimum swaps
// to group all 1's together
function minSwaps($arr, $n)
{
$numberOfOnes = 0;
// find total number of
// all 1's in the array
for ($i = 0; $i < $n; $i++)
{
if ($arr[$i] == 1)
$numberOfOnes++;
}
// length of subarray to check for
$x = $numberOfOnes;
$count_ones = 0;
$maxOnes;
// Find 1's for first
// subarray of length x
for($i = 0; $i < $x; $i++)
{
if($arr[$i] == 1)
$count_ones++;
}
$maxOnes = $count_ones;
// using sliding window
// technique to find
// max number of ones in
// subarray of length x
for ($i = 1; $i <= $n - $x; $i++)
{
// first remove leading
// element and check
// if it is equal to 1
// then decrement the
// value of count_ones by 1
if ($arr[$i - 1] === 1)
$count_ones--;
// Now add trailing
// element and check
// if it is equal to 1
// Then increment
// the value of count_ones by 1
if($arr[$i + $x - 1] === 1)
$count_ones++;
if ($maxOnes < $count_ones)
$maxOnes = $count_ones;
}
// calculate number of zeros in subarray
// of length x with maximum number of 1's
$numberOfZeroes = $x - $maxOnes;
return $numberOfZeroes;
}
// Driver Code
$a = array(0, 0, 1, 0, 1, 1, 0, 0, 1);
$n = 9;
echo minSwaps($a, $n);
// This code is contributed by Anuj_67
?>
java 描述语言
<script>
// Javascript code for minimum swaps
// required to group all 1's together
// Function to find minimum swaps
// to group all 1's together
function minSwaps(arr, n)
{
let numberOfOnes = 0;
// find total number of all 1's in the array
for (let i = 0; i < n; i++) {
if (arr[i] == 1)
numberOfOnes++;
}
// length of subarray to check for
let x = numberOfOnes;
let count_ones = 0, maxOnes;
// Find 1's for first subarray of length x
for(let i = 0; i < x; i++){
if(arr[i] == 1)
count_ones++;
}
maxOnes = count_ones;
// using sliding window technique to find
// max number of ones in subarray of length x
for (let i = 1; i <= n-x; i++) {
// first remove leading element and check
// if it is equal to 1 then decrement the
// value of count_ones by 1
if (arr[i - 1] == 1)
count_ones--;
// Now add trailing element and check
// if it is equal to 1 Then increment
// the value of count_ones by 1
if(arr[i + x - 1] == 1)
count_ones++;
if (maxOnes < count_ones)
maxOnes = count_ones;
}
// calculate number of zeros in subarray
// of length x with maximum number of 1's
let numberOfZeroes = x - maxOnes;
return numberOfZeroes;
}
let a = [0, 0, 1, 0, 1, 1, 0, 0, 1];
let n = a.length;
document.write(minSwaps(a, n));
</script>
Output
1
时间复杂度: O(n) 辅助空间: O(1) 感谢 杰拉先生 提出这个方法。
滑动窗口易懂版
算法:
1.将总数存储在一个变量中，比如计数。这将是窗口大小。
2.初始化一个变量来存储大小计数的所有子数组中的最大数量的 1，并初始化一个变量来存储当前窗口中的 1 计数。
3.现在迭代数组，一旦你达到窗口大小，将该窗口中的“1”的数量与目前所有窗口中的“1”的最大数量进行比较，如果当前窗口中的“1”数量更多，则更新“1”的最大数量。如果窗口的第一个元素是 1，则减少当前计数。
4.答案将是 1 的总数–所有窗口中 1 的最大数量。
C++
#include <bits/stdc++.h>
using namespace std;
int minSwaps(int arr[], int n)
{
int totalCount = 0; // To store total number of ones
// count total no of ones
for (int i = 0; i < n; i++)
totalCount += arr[i];
int currCount = 0; // To store count of ones in current window
int maxCount = 0; // To store maximum count ones out of all windows
int i = 0; // start of window
int j = 0; // end of window
while (j < n)
{
currCount += arr[j];
// update maxCount when reach window size i.e. total count of ones in array
if ((j - i + 1) == totalCount)
{
maxCount = max(maxCount, currCount);
if (arr[i] == 1)
currCount--; // decrease current count if first element of window is 1
i++; // slide window
}
j++;
}
return totalCount - maxCount; // return total no of ones in array - maximum count of ones out of all windows
}
// Driver Code
int main()
{
int a[] = {1, 0, 1, 0, 1, 1};
int n = sizeof(a) / sizeof(a[0]);
cout << minSwaps(a, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
import java.io.*;
import java.util.*;
class GFG{
static int minSwaps(int[] arr, int n)
{
// To store total number of ones
int totalCount = 0;
// Count total no of ones
int i;
for(i = 0; i < n; i++)
totalCount += arr[i];
int currCount = 0; // To store count of ones in current window
int maxCount = 0; // To store maximum count ones out
// of all windows
// start of window
i = 0;
// end of window
int j = 0;
while (j < n)
{
currCount += arr[j];
// update maxCount when reach window size i.e.
// total count of ones in array
if ((j - i + 1) == totalCount)
{
maxCount = Math.max(maxCount, currCount);
if (arr[i] == 1)
currCount--; // decrease current count
// if first element of
// window is 1
// slide window
i++;
}
j++;
}
return totalCount - maxCount; // return total no of ones in array
// - maximum count of ones out of
// all windows
}
// Driver Code
public static void main(String args[])
{
int[] a = { 1, 0, 1, 0, 1, 1 };
int n = a.length;
System.out.println(minSwaps(a, n));
}
}
// This code is contributed by shivanisinghss2110
计算机编程语言
def minSwaps(arr, n):
# To store total number of ones
totalCount = 0
# count total no of ones
for i in range(0,n):
totalCount += arr[i]
currCount = 0 # To store count of ones in current window
maxCount = 0 # To store maximum count ones out of all windows
i = 0 # start of window
j = 0 # end of window
while (j < n):
currCount += arr[j]
# update maxCount when reach window size i.e. total count of ones in array
if ((j - i + 1) == totalCount):
maxCount = max(maxCount, currCount)
if (arr[i] == 1):
currCount -= 1
# decrease current count if first element of window is 1
i += 1 # slide window
j += 1
return totalCount - maxCount # return total no of ones in array - maximum count of ones out of all windows
# Driver Code
a = [1, 0, 1, 0, 1, 1]
n = len(a)
print(minSwaps(a, n))
# this code is contributed by shivanisighss2110
C
using System;
class GFG{
static int minSwaps(int[] arr, int n)
{
// To store total number of ones
int totalCount = 0;
// Count total no of ones
int i;
for(i = 0; i < n; i++)
totalCount += arr[i];
int currCount = 0; // To store count of ones in current window
int maxCount = 0; // To store maximum count ones out
// of all windows
// start of window
i = 0;
// end of window
int j = 0;
while (j < n)
{
currCount += arr[j];
// update maxCount when reach window size i.e.
// total count of ones in array
if ((j - i + 1) == totalCount)
{
maxCount = Math.Max(maxCount, currCount);
if (arr[i] == 1)
currCount--; // decrease current count
// if first element of
// window is 1
// slide window
i++;
}
j++;
}
return totalCount - maxCount; // return total no of ones in array
// - maximum count of ones out of
// all windows
}
// Driver Code
public static void Main()
{
int[] a = { 1, 0, 1, 0, 1, 1 };
int n = a.Length;
Console.WriteLine(minSwaps(a, n));
}
}
// This code is contributed by ukasp
java 描述语言
<script>
function minSwaps(arr, n)
{
// To store total number of ones
let totalCount = 0;
// Count total no of ones
let i;
for(i = 0; i < n; i++)
totalCount += arr[i];
let currCount = 0; // To store count of ones in current window
let maxCount = 0; // To store maximum count ones out
// of all windows
// start of window
i = 0;
// end of window
let j = 0;
while (j < n)
{
currCount += arr[j];
// update maxCount when reach window size i.e.
// total count of ones in array
if ((j - i + 1) == totalCount)
{
maxCount = Math.max(maxCount, currCount);
if (arr[i] == 1)
currCount--; // decrease current count
// if first element of
// window is 1
// slide window
i++;
}
j++;
}
return totalCount - maxCount; // return total no of ones in array
// - maximum count of ones out of
// all windows
}
// Driver Code
let a = [ 1, 0, 1, 0, 1, 1 ];
let n = a.length;
document.write(minSwaps(a, n));
// This code is contributed by shivanisinghss2110
</script>
Output
1
复杂性:
时间:O(n)
空间:0(1)
版权属于：月萌API www.moonapi.com，转载请注明出处
